Electrochemistry

The oxidation state is a formalism that assigns a charge to an atom in a compound based on electronegativity. The rules for assigning oxidation states:

1. The oxidation state of an uncombined element is zero.
2. The sum of oxidation states in a neutral compound is zero; in a polyatomic ion, it equals the ion charge.
3. Group 1 metals: $+1$. Group 2 metals: $+2$. Al: $+3$.
4. Hydrogen is $+1$ (except in metal hydrides, where it is $-1$).
5. Oxygen is $-2$ (except in peroxides, where it is $-1$And in $\mathrm{OF}_2$Where it is $+2$).
6. Fluorine is always $-1$ in compounds.
7. Halogens are $-1$Except when bonded to more electronegative elements or in polyatomic ions.

Worked Example. Determine the oxidation state of $\mathrm{Cr}$ in $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$.

$$2(+1) + 2x + 7(-2) = 0$$ $$2 + 2x - 14 = 0$$ $$2x = 12 \implies x = +6$$

Redox reactions are split into two half-equations:

• Oxidation: loss of electrons (increase in oxidation state).
• Reduction: gain of electrons (decrease in oxidation state).

The mnemonic OIL RIG: Oxidation Is Loss, Reduction Is Gain.

1. Balance all atoms except $\mathrm{O}$ and $\mathrm{H}$.
2. Balance $\mathrm{O}$ by adding $\mathrm{H}_2\mathrm{O}$.
3. Balance $\mathrm{H}$ by adding $\mathrm{H}^+$.
4. Balance charge by adding electrons ($e^-$).

Example. Balance $\mathrm{MnO}_4^- \to \mathrm{Mn}^{2+}$ in acidic solution.

1. $\mathrm{Mn}$ already balanced.
2. Add $\mathrm{H}_2\mathrm{O}$: $\mathrm{MnO}_4^- \to \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}$.
3. Add $\mathrm{H}^+$: $\mathrm{MnO}_4^- + 8\mathrm{H}^+ \to \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}$.
4. Balance charge: LHS = $-1 + 8 = +7$; RHS = $+2$. Add $5e^-$ to LHS: $\mathrm{MnO}_4^- + 8\mathrm{H}^+ + 5e^- \to \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}$.

Multiply each half-equation by an appropriate factor so that the electrons cancel, then add.

Example. Combine the above with $\mathrm{Fe}^{2+} \to \mathrm{Fe}^{3+} + e^-$:

$$\mathrm{MnO}_4^- + 8\mathrm{H}^+ + 5e^- \to \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}$$ $$5\mathrm{Fe}^{2+} \to 5\mathrm{Fe}^{3+} + 5e^-$$ $$\mathrm{MnO}_4^- + 8\mathrm{H}^+ + 5\mathrm{Fe}^{2+} \to \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O} + 5\mathrm{Fe}^{3+}$$

The standard electrode potential $E^\circ$ is the potential difference (voltage) of a half-cell relative to the standard hydrogen electrode under standard conditions (298\,\mathrm{K}$$100\,\mathrm{kPa}$$1\,\mathrm{mol/dm}^3 solutions).

The SHE is the reference electrode:

$$2\mathrm{H}^+(aq, 1\,\mathrm{mol/dm}^3) + 2e^- \rightleftharpoons \mathrm{H}_2(g, 100\,\mathrm{kPa})$$

By definition, $E^\circ(\mathrm{H}^+/\mathrm{H}_2) = 0.00\,\mathrm{V}$.

To measure the $E^\circ$ of a half-cell, it is connected to the SHE, and the cell EMF is measured. A salt bridge ( $\mathrm{KNO}_3$ solution) maintains electrical neutrality.

Half-equation	$E^\circ$ (V)
$\mathrm{F}_2 + 2e^- \rightleftharpoons 2\mathrm{F}^-$	$+2.87$
$\mathrm{MnO}_4^- + 8\mathrm{H}^+ + 5e^- \rightleftharpoons \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}$	$+1.51$
$\mathrm{Cl}_2 + 2e^- \rightleftharpoons 2\mathrm{Cl}^-$	$+1.36$
$\mathrm{Ag}^+ + e^- \rightleftharpoons \mathrm{Ag}$	$+0.80$
$\mathrm{Cu}^{2+} + 2e^- \rightleftharpoons \mathrm{Cu}$	$+0.34$
$2\mathrm{H}^+ + 2e^- \rightleftharpoons \mathrm{H}_2$	$0.00$
$\mathrm{Fe}^{2+} + 2e^- \rightleftharpoons \mathrm{Fe}$	$-0.44$
$\mathrm{Zn}^{2+} + 2e^- \rightleftharpoons \mathrm{Zn}$	$-0.76$
$\mathrm{Na}^+ + e^- \rightleftharpoons \mathrm{Na}$	$-2.71$
$\mathrm{Li}^+ + e^- \rightleftharpoons \mathrm{Li}$	$-3.04$

Key observations:

• More positive $E^\circ$: species is a stronger oxidising agent (more readily reduced).
• More negative $E^\circ$: species is a stronger reducing agent (more readily oxidised).
• A species can oxidise any species below it in the series (with more negative $E^\circ$).

For an electrochemical cell consisting of two half-cells:

$$E^\circ_\mathrm{cell} = E^\circ_\mathrm{reduction} - E^\circ_\mathrm{oxidation}$$

Or equivalently:

$$E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode}$$

Where the cathode is the electrode where reduction occurs and the anode is where oxidation occurs.

Worked Example. Calculate the standard cell EMF for a $\mathrm{Zn}/\mathrm{Cu}$ cell.

$$\mathrm{Zn}^{2+} + 2e^- \rightleftharpoons \mathrm{Zn} \quad E^\circ = -0.76\,\mathrm{V}$$ $$\mathrm{Cu}^{2+} + 2e^- \rightleftharpoons \mathrm{Cu} \quad E^\circ = +0.34\,\mathrm{V}$$

$\mathrm{Zn}$ is oxidised (anode), $\mathrm{Cu}^{2+}$ is reduced (cathode).

$$E^\circ_\mathrm{cell} = 0.34 - (-0.76) = +1.10\,\mathrm{V}$$

A redox reaction is thermodynamically feasible if $E^\circ_\mathrm{cell} \gt 0$. This is equivalent to $\Delta G^\circ \lt 0$:

$$\Delta G^\circ = -nFE^\circ_\mathrm{cell}$$

Where $n$ is the number of moles of electrons transferred and $F$ is the Faraday constant ($96485\,\mathrm{C/mol}$).

Caveats:

• $E^\circ_\mathrm{cell} \gt 0$ indicates thermodynamic feasibility, not necessarily a fast reaction (kinetic barriers may exist).
• Non-standard conditions (concentrations other than $1\,\mathrm{mol/dm}^3$) change the actual cell EMF via the Nernst equation (beyond A-Level but useful to note).

The Nernst equation relates the cell potential under non-standard conditions to the standard cell potential:

$$E = E^\circ - \frac{RT}{nF}\ln Q$$

Where $Q$ is the reaction quotient (same form as the equilibrium expression but with initial concentrations).

At $298\,\mathrm{K}$Substituting $R = 8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}$ and $F = 96485\,\mathrm{C/mol}$:

$$E = E^\circ - \frac{0.0257}{n}\ln Q = E^\circ - \frac{0.0592}{n}\log_{10} Q$$

Worked Example. Calculate the cell EMF for a $\mathrm{Zn}/\mathrm{Cu}$ cell where $[\mathrm{Zn}^{2+}] = 0.010\,\mathrm{mol/dm}^3$ and $[\mathrm{Cu}^{2+}] = 2.0\,\mathrm{mol/dm}^3$ at $298\,\mathrm{K}$.

$E^\circ_\mathrm{cell} = +1.10\,\mathrm{V}$ (from above). $n = 2$.

$$Q = \frac{[\mathrm{Zn}^{2+}]}{[\mathrm{Cu}^{2+}]} = \frac{0.010}{2.0} = 0.005$$ $$E = 1.10 - \frac{0.0592}{2}\log_{10}(0.005) = 1.10 - 0.0296 \times (-2.30) = 1.10 + 0.068 = 1.17\,\mathrm{V}$$

The non-standard cell EMF is higher than the standard value because the product ion ($\mathrm{Zn}^{2+}$) concentration is lower and the reactant ion ($\mathrm{Cu}^{2+}$) concentration is higher than standard conditions, driving the reaction further to the right.

A concentration cell consists of two half-cells with the same redox couple but different concentrations. The cell EMF arises purely from the concentration difference:

$$E_\mathrm{cell} = -\frac{0.0592}{n}\log_{10}\frac{[\mathrm{M}^{n+}]_\mathrm{dilute}}{[\mathrm{M}^{n+}]_\mathrm{concentrated}}$$

Example: A $\mathrm{Cu}/\mathrm{Cu}^{2+}$ concentration cell with $[\mathrm{Cu}^{2+}]_\mathrm{left} = 0.001\,\mathrm{mol/dm}^3$ and $[\mathrm{Cu}^{2+}]_\mathrm{right} = 1.0\,\mathrm{mol/dm}^3$:

$$E = -\frac{0.0592}{2}\log_{10}\frac{0.001}{1.0} = -0.0296 \times (-3) = +0.089\,\mathrm{V}$$

The dilute half-cell acts as the anode (oxidation), and the concentrated half-cell acts as the cathode (reduction). Equilibrium is reached when the concentrations equalise.

Cells are written as:

$$\mathrm{Anode (oxidation)} \parallel \mathrm{Cathode (reduction)}$$

Example: $\mathrm{Zn}(s) \mid \mathrm{Zn}^{2+}(aq) \parallel \mathrm{Cu}^{2+}(aq) \mid \mathrm{Cu}(s)$

The single vertical line represents a phase boundary. The double vertical line represents the salt bridge. The anode (oxidation) is written on the left; the cathode (reduction) on the right.

A fuel cell converts the chemical energy of a fuel ( $\mathrm{H}_2$) directly into electrical energy, continuously supplied from an external source.

Alkaline conditions:

Anode (oxidation): $2\mathrm{H}_2 + 4\mathrm{OH}^- \to 4\mathrm{H}_2\mathrm{O} + 4e^-$

Cathode (reduction): $\mathrm{O}_2 + 2\mathrm{H}_2\mathrm{O} + 4e^- \to 4\mathrm{OH}^-$

Overall: $2\mathrm{H}_2 + \mathrm{O}_2 \to 2\mathrm{H}_2\mathrm{O}$

Acidic conditions:

Anode: $2\mathrm{H}_2 \to 4\mathrm{H}^+ + 4e^-$

Cathode: $\mathrm{O}_2 + 4\mathrm{H}^+ + 4e^- \to 2\mathrm{H}_2\mathrm{O}$

Overall: $2\mathrm{H}_2 + \mathrm{O}_2 \to 2\mathrm{H}_2\mathrm{O}$

Advantages: Higher efficiency than combustion; no pollution (only water as product); quiet operation.

Disadvantages: Hydrogen storage and transport challenges; expensive catalysts (platinum); limited infrastructure.

Efficiency calculation: The maximum theoretical efficiency of a fuel cell is:

$$\text{Efficiency} = \frac{\Delta G^\circ}{\Delta H^\circ} \times 100$$

For the $\mathrm{H}_2/\mathrm{O}_2$ fuel cell at $298\,\mathrm{K}$:

$$\Delta G^\circ = -2 \times 237 = -474\,\mathrm{kJ/mol}, \quad \Delta H^\circ = -2 \times 286 = -572\,\mathrm{kJ/mol}$$ $$\text{Efficiency} = \frac{474}{572} \times 100 = 83\%$$

In practice, efficiencies of 40—60% are achieved due to overpotentials (additional voltage required beyond the theoretical value), internal resistance, and fuel crossover.

Overpotential: The actual operating voltage of a fuel cell is less than the theoretical $E^\circ_\mathrm{cell} = 1.23\,\mathrm{V}$ for $\mathrm{H}_2/\mathrm{O}_2$. Typical operating voltages are $0.6$—$0.8\,\mathrm{V}$. The difference is the overpotential, caused by kinetic barriers at the electrodes (especially the oxygen reduction reaction at the cathode, which is inherently slow).

Type	Fuel	Electrolyte	Operating temp	Application
PEM (proton exchange membrane)	$\mathrm{H}_2$	Solid polymer	80$^\circ$C	Vehicles
Alkaline (AFC)	$\mathrm{H}_2$	KOH solution	60—200$^\circ$C	Space (Apollo)
Solid oxide (SOFC)	$\mathrm{H}_2$CO, $\mathrm{CH}_4$	Yttria-stabilised zirconia	500—1000$^\circ$C	Power stations
Molten carbonate (MCFC)	$\mathrm{H}_2$CO, $\mathrm{CH}_4$	$\mathrm{Li}_2\mathrm{CO}_3/\mathrm{K}_2\mathrm{CO}_3$	650$^\circ$C	Power stations

Solid oxide fuel cells can use carbon monoxide and methane directly as fuels (internal reforming), and their high operating temperature makes them suitable for combined heat and power systems.

Lead-acid battery:

Anode: $\mathrm{Pb}(s) + \mathrm{SO}_4^{2-}(aq) \to \mathrm{PbSO}_4(s) + 2e^-$

Cathode: $\mathrm{PbO}_2(s) + \mathrm{SO}_4^{2-}(aq) + 4\mathrm{H}^+(aq) + 2e^- \to \mathrm{PbSO}_4(s) + 2\mathrm{H}_2\mathrm{O}(l)$

Overall: $\mathrm{Pb} + \mathrm{PbO}_2 + 2\mathrm{H}_2\mathrm{SO}_4 \to 2\mathrm{PbSO}_4 + 2\mathrm{H}_2\mathrm{O}$

EMF $\approx 2\,\mathrm{V}$ per cell; six cells in series give $12\,\mathrm{V}$ car battery.

Lithium-ion battery:

Anode (oxidation): $\mathrm{LiCoO}_2 \to \mathrm{Li}_{1-x}\mathrm{CoO}_2 + x\mathrm{Li}^+ + xe^-$

Cathode (reduction): $\mathrm{C}_6 + x\mathrm{Li}^+ + xe^- \to \mathrm{Li}_x\mathrm{C}_6$

Lithium ions migrate between electrodes through an electrolyte. EMF $\approx 3.7\,\mathrm{V}$.

Electrolysis uses electrical energy to drive a non-spontaneous redox reaction. The electrolyte is an ionic compound, molten or in solution, that conducts electricity through the movement of ions.

• Anode (+): oxidation occurs; anions are attracted.
• Cathode (-): reduction occurs; cations are attracted.

In aqueous solutions, both the cation and $\mathrm{H}^+$ (from water) can be reduced at the cathode, and both the anion and $\mathrm{OH}^-$ (from water) can be oxidised at the anion. The species that is more reduced/oxidised (higher/lower $E^\circ$) is preferentially discharged.

At the cathode (reduction):

If the metal is below hydrogen in the reactivity series (e.g. \mathrm{Cu}$$\mathrm{Ag}$$\mathrm{Au}), the metal is deposited. If the metal is more reactive (e.g. \mathrm{Na}$$\mathrm{K}$$\mathrm{Mg}), hydrogen is evolved:

$$2\mathrm{H}_2\mathrm{O}(l) + 2e^- \to \mathrm{H}_2(g) + 2\mathrm{OH}^-(aq)$$

At the anode (oxidation):

If the anion is a halide (\mathrm{Cl}^-$$\mathrm{Br}^-$$\mathrm{I}^-), the halogen is produced. For all other anions (including \mathrm{SO}_4^{2-}$$\mathrm{NO}_3^-), oxygen is evolved from water:

$$4\mathrm{OH}^-(aq) \to \mathrm{O}_2(g) + 2\mathrm{H}_2\mathrm{O}(l) + 4e^-$$

Or equivalently:

$$2\mathrm{H}_2\mathrm{O}(l) \to \mathrm{O}_2(g) + 4\mathrm{H}^+(aq) + 4e^-$$

First Law: The mass of substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed.

$$M = \frac{Q \cdot M}{n \cdot F}$$

Second Law: When the same quantity of electricity is passed through different electrolytes, the masses of substances deposited are proportional to their equivalent masses ($M/n$Where $n$ is the number of electrons transferred per ion).

$$Q = I \times t$$

Where $Q$ is charge in Coulombs, $I$ is current in Amperes, and $t$ is time in seconds.

$F = 96485\,\mathrm{C/mol}$ (Faraday constant) — the charge carried by one mole of electrons.

Worked Example. Calculate the mass of copper deposited when a current of $2.50\,\mathrm{A}$ is passed through $\mathrm{CuSO}_4$ solution for $30.0$ minutes.

$$Q = 2.50 \times 30.0 \times 60 = 4500\,\mathrm{C}$$ $$N(e^-) = \frac{4500}{96485} = 0.0466\,\mathrm{mol}$$ $$\mathrm{Cu}^{2+} + 2e^- \to \mathrm{Cu}$$ $$N(\mathrm{Cu}) = \frac{0.0466}{2} = 0.0233\,\mathrm{mol}$$ $$M(\mathrm{Cu}) = 0.0233 \times 63.5 = 1.48\,\mathrm{g}$$

Alternatively, using the formula directly:

$$M = \frac{Q \cdot M}{n \cdot F} = \frac{4500 \times 63.5}{2 \times 96485} = \frac{285750}{192970} = 1.48\,\mathrm{g}$$

Rusting is an electrochemical process requiring both water and oxygen:

Anode (oxidation): $\mathrm{Fe}(s) \to \mathrm{Fe}^{2+}(aq) + 2e^-$

Cathode (reduction): $\mathrm{O}_2(g) + 2\mathrm{H}_2\mathrm{O}(l) + 4e^- \to 4\mathrm{OH}^-(aq)$

The $\mathrm{Fe}^{2+}$ is further oxidised and reacts with $\mathrm{OH}^-$ to form hydrated iron(III) oxide (rust).

Method	Principle
Painting / oiling	Physical barrier to water and oxygen
Galvanising (zinc coating)	Zinc acts as a sacrificial anode (more reactive than iron); even if scratched, zinc corrodes preferentially
Tin plating	Physical barrier; if scratched, tin accelerates rusting (tin is less reactive than iron)
Sacrificial protection	Attaching blocks of a more reactive metal (e.g. \mathrm{Mg}$$\mathrm{Zn}) to the iron structure
Alloying (stainless steel)	Chromium forms a protective oxide layer

1. Incorrect sign in the cell EMF formula. $E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode}$. Subtracting in the wrong order gives a negative value.

2. Confusing oxidation and reduction at the electrodes. In electrolysis, the anode is positive (oxidation), cathode is negative (reduction). In a galvanic cell, the anode is negative (oxidation), cathode is positive (reduction). The signs are reversed.

3. Forgetting to halve the Faraday calculation for divalent ions. For $\mathrm{Cu}^{2+}$Two moles of electrons deposit one mole of copper.

4. Incorrect half-equations for aqueous electrolysis. Students often write the anion oxidation instead of water oxidation when the anion is not a halide.

5. Stating that a reaction “will occur” based only on $E^\circ_\mathrm{cell} \gt 0$. Thermodynamic feasibility does not guarantee kinetic viability. The reaction may be too slow to observe.

6. Confusing $E^\circ_\mathrm{cell}$ with $E^\circ$ of individual half-cells. $E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode}$. Always subtract the anode potential from the cathode potential.

7. Forgetting that the SHE is defined at specific conditions. The standard hydrogen electrode operates at 298\,\mathrm{K}$$100\,\mathrm{kPa} $\mathrm{H}_2$ pressure, and $1.0\,\mathrm{mol/dm}^3$ $\mathrm{H}^+$ concentration. Any deviation requires the Nernst equation.

A redox reaction is thermodynamically feasible if $E^\circ_\mathrm{cell} \gt 0$ (i.e. $\Delta G^\circ \lt 0$). However:

1. A positive $E^\circ_\mathrm{cell}$ indicates thermodynamic feasibility, not kinetic likelihood. Some reactions with positive $E^\circ_\mathrm{cell}$ are extremely slow (e.g. $\mathrm{H}_2$ + $\mathrm{O}_2$ at room temperature).
2. The reaction may require a catalyst, activation energy, or specific conditions to proceed at an appreciable rate.
3. Non-standard conditions can reverse the feasibility (use the Nernst equation).

Disproportionation: A species in an intermediate oxidation state simultaneously oxidises and reduces itself:

$$2\mathrm{Cu}^+(aq) \to \mathrm{Cu}(s) + \mathrm{Cu}^{2+}(aq)$$

Check feasibility: $E^\circ_\mathrm{cell} = E^\circ(\mathrm{Cu}^+/\mathrm{Cu}) - E^\circ(\mathrm{Cu}^{2+}/\mathrm{Cu}^+) = 0.52 - 0.15 = +0.37\,\mathrm{V}$. Spontaneous. $\mathrm{Cu}^+$ disproportionates in aqueous solution.

Comproportionation: The reverse of disproportionation. Two species in different oxidation states react to form a species in an intermediate state:

$$\mathrm{Fe}(s) + 2\mathrm{Fe}^{3+}(aq) \to 3\mathrm{Fe}^{2+}(aq)$$

Half-equation	$E^\circ$ (V)
$\mathrm{F}_2 + 2e^- \rightleftharpoons 2\mathrm{F}^-$	$+2.87$
$\mathrm{MnO}_4^- + 8\mathrm{H}^+ + 5e^- \rightleftharpoons \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}$	$+1.51$
$\mathrm{Cl}_2 + 2e^- \rightleftharpoons 2\mathrm{Cl}^-$	$+1.36$
$\mathrm{Cr}_2\mathrm{O}_7^{2-} + 14\mathrm{H}^+ + 6e^- \rightleftharpoons 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O}$	$+1.33$
$\mathrm{O}_2 + 4\mathrm{H}^+ + 4e^- \rightleftharpoons 2\mathrm{H}_2\mathrm{O}$	$+1.23$
$\mathrm{Br}_2 + 2e^- \rightleftharpoons 2\mathrm{Br}^-$	$+1.07$
$\mathrm{NO}_3^- + 4\mathrm{H}^+ + 3e^- \rightleftharpoons \mathrm{NO} + 2\mathrm{H}_2\mathrm{O}$	$+0.96$
$\mathrm{Ag}^+ + e^- \rightleftharpoons \mathrm{Ag}$	$+0.80$
$\mathrm{Fe}^{3+} + e^- \rightleftharpoons \mathrm{Fe}^{2+}$	$+0.77$
$\mathrm{I}_2 + 2e^- \rightleftharpoons 2\mathrm{I}^-$	$+0.54$
$\mathrm{Cu}^+ + e^- \rightleftharpoons \mathrm{Cu}$	$+0.52$
$\mathrm{O}_2 + 2\mathrm{H}_2\mathrm{O} + 4e^- \rightleftharpoons 4\mathrm{OH}^-$	$+0.40$
$\mathrm{Cu}^{2+} + 2e^- \rightleftharpoons \mathrm{Cu}$	$+0.34$
$2\mathrm{H}^+ + 2e^- \rightleftharpoons \mathrm{H}_2$	$0.00$ (by definition)
$\mathrm{Pb}^{2+} + 2e^- \rightleftharpoons \mathrm{Pb}$	$-0.13$
$\mathrm{Sn}^{2+} + 2e^- \rightleftharpoons \mathrm{Sn}$	$-0.14$
$\mathrm{Ni}^{2+} + 2e^- \rightleftharpoons \mathrm{Ni}$	$-0.25$
$\mathrm{Fe}^{2+} + 2e^- \rightleftharpoons \mathrm{Fe}$	$-0.44$
$\mathrm{Zn}^{2+} + 2e^- \rightleftharpoons \mathrm{Zn}$	$-0.76$
$\mathrm{Al}^{3+} + 3e^- \rightleftharpoons \mathrm{Al}$	$-1.66$
$\mathrm{Mg}^{2+} + 2e^- \rightleftharpoons \mathrm{Mg}$	$-2.37$
$\mathrm{Na}^+ + e^- \rightleftharpoons \mathrm{Na}$	$-2.71$
$\mathrm{Ca}^{2+} + 2e^- \rightleftharpoons \mathrm{Ca}$	$-2.87$
$\mathrm{K}^+ + e^- \rightleftharpoons \mathrm{K}$	$-2.93$
$\mathrm{Li}^+ + e^- \rightleftharpoons \mathrm{Li}$	$-3.04$

• Ore: Bauxite ($\mathrm{Al}_2\mathrm{O}_3$), purified to alumina.
• Electrolyte: Molten cryolite ($\mathrm{Na}_3\mathrm{AlF}_6$) at approximately $950^\circ\mathrm{C}$. Cryolite lowers the melting point of $\mathrm{Al}_2\mathrm{O}_3$ from $2072^\circ\mathrm{C}$ to approximately $950^\circ\mathrm{C}$Making the process economically viable.
• Electrodes: Carbon (graphite) anode and cathode.

Cathode: $\mathrm{Al}^{3+} + 3e^- \to \mathrm{Al}(l)$

Anode: $2\mathrm{O}^{2-} \to \mathrm{O}_2 + 4e^-$

The oxygen reacts with the carbon anode: $\mathrm{C} + \mathrm{O}_2 \to \mathrm{CO}_2$. The anode is consumed and must be regularly replaced.

Electroplating deposits a thin layer of a metal onto a conductive surface. The object to be plated is the cathode, the plating metal is the anode, and the electrolyte contains ions of the plating metal.

Example: Copper plating of a steel spoon.

• Cathode (spoon): $\mathrm{Cu}^{2+}(aq) + 2e^- \to \mathrm{Cu}(s)$
• Anode (copper): $\mathrm{Cu}(s) \to \mathrm{Cu}^{2+}(aq) + 2e^-$
• Electrolyte: $\mathrm{CuSO}_4(aq)$

The concentration of $\mathrm{Cu}^{2+}$ in solution remains constant because the anode dissolves at the same rate as the cathode deposits copper.

Problem 1

Use standard electrode potentials to determine whether $\mathrm{Br}_2$ can oxidise $\mathrm{Fe}^{2+}$ to $\mathrm{Fe}^{3+}$.

Given: $\mathrm{Br}_2 + 2e^- \rightleftharpoons 2\mathrm{Br}^-$ ($E^\circ = +1.07\,\mathrm{V}$), $\mathrm{Fe}^{3+} + e^- \rightleftharpoons \mathrm{Fe}^{2+}$ ($E^\circ = +0.77\,\mathrm{V}$).

Solution:

$\mathrm{Br}_2$ would be reduced (cathode), $\mathrm{Fe}^{2+}$ would be oxidised (anode).

$$E^\circ_\mathrm{cell} = 1.07 - 0.77 = +0.30\,\mathrm{V}$$

Since $E^\circ_\mathrm{cell} \gt 0$The reaction is thermodynamically feasible.

Problem 2

How long must a current of $5.00\,\mathrm{A}$ pass through molten $\mathrm{Al}_2\mathrm{O}_3$ to produce $10.0\,\mathrm{g}$ of aluminium?

Solution:

$$\mathrm{Al}^{3+} + 3e^- \to \mathrm{Al}$$ $$N(\mathrm{Al}) = \frac{10.0}{27.0} = 0.370\,\mathrm{mol}$$ $$N(e^-) = 3 \times 0.370 = 1.111\,\mathrm{mol}$$ $$Q = 1.111 \times 96485 = 107,200\,\mathrm{C}$$ $$T = \frac{Q}{I} = \frac{107200}{5.00} = 21440\,\mathrm{s} = 357\,\mathrm{minutes} = 5.96\,\mathrm{hours}$$

Problem 3

A voltaic cell is constructed from a $\mathrm{Zn}(s) \mid \mathrm{Zn}^{2+}(aq)$ half-cell and an $\mathrm{Ag}(s) \mid \mathrm{Ag}^+(aq)$ half-cell. Calculate the standard cell EMF, write the conventional cell representation, and determine $\Delta G^\circ$.

Given: E^\circ(\mathrm{Zn}^{2+}/\mathrm{Zn}) = -0.76\,\mathrm{V}$$E^\circ(\mathrm{Ag}^+/\mathrm{Ag}) = +0.80\,\mathrm{V}.

Solution:

$\mathrm{Zn}$ has the more negative $E^\circ$So it is oxidised (anode). $\mathrm{Ag}^+$ is reduced (cathode).

$E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode} = 0.80 - (-0.76) = +1.56\,\mathrm{V}$

Conventional cell representation:

$$\mathrm{Zn}(s) \mid \mathrm{Zn}^{2+}(aq) \parallel \mathrm{Ag}^+(aq) \mid \mathrm{Ag}(s)$$

Half-equations:

Anode: $\mathrm{Zn}(s) \to \mathrm{Zn}^{2+}(aq) + 2e^-$

Cathode: $\mathrm{Ag}^+(aq) + e^- \to \mathrm{Ag}(s)$

Overall (multiplying the cathode equation by 2): $\mathrm{Zn}(s) + 2\mathrm{Ag}^+(aq) \to \mathrm{Zn}^{2+}(aq) + 2\mathrm{Ag}(s)$

$n = 2$ electrons transferred.

$$\Delta G^\circ = -nFE^\circ_\mathrm{cell} = -2 \times 96485 \times 1.56 = -301,000\,\mathrm{J/mol} = -301\,\mathrm{kJ/mol}$$

The large negative $\Delta G^\circ$ confirms that the reaction is strongly spontaneous under standard conditions.

Problem 4

In the electrolysis of aqueous $\mathrm{CuSO}_4$ using inert platinum electrodes: (a) Identify the products at each electrode. (b) Write half-equations. (c) Calculate the volume of gas produced at the anode when a current of $0.50\,\mathrm{A}$ passes for 1 hour at $298\,\mathrm{K}$ and $100\,\mathrm{kPa}$.

Solution:

(a) At the cathode: $\mathrm{Cu}^{2+}$ is below hydrogen in the reactivity series, so copper is deposited (not hydrogen). At the anode: $\mathrm{SO}_4^{2-}$ is not a halide, so oxygen is evolved from water.

(b) Cathode: $\mathrm{Cu}^{2+}(aq) + 2e^- \to \mathrm{Cu}(s)$

Anode: $4\mathrm{OH}^-(aq) \to \mathrm{O}_2(g) + 2\mathrm{H}_2\mathrm{O}(l) + 4e^-$

(c) $Q = 0.50 \times 3600 = 1800\,\mathrm{C}$

$n(e^-) = 1800 / 96485 = 0.01866\,\mathrm{mol}$

From the anode half-equation, 4 moles of electrons produce 1 mole of $\mathrm{O}_2$:

$n(\mathrm{O}_2) = 0.01866 / 4 = 0.00467\,\mathrm{mol}$

Using $pV = nRT$:

$$V = \frac{nRT}{p} = \frac{0.00467 \times 8.314 \times 298}{100 \times 10^3} = \frac{11.57}{100000} = 1.16 \times 10^{-4}\,\mathrm{m}^3 = 0.116\,\mathrm{dm}^3 = 116\,\mathrm{cm}^3$$

Problem 5

A concentration cell is constructed with two $\mathrm{Cu}$ electrodes. One half-cell contains $\mathrm{Cu}^{2+}$ at $1.00\,\mathrm{mol/dm}^3$ and the other contains $\mathrm{Cu}^{2+}$ at $0.00100\,\mathrm{mol/dm}^3$ at $298\,\mathrm{K}$.

(a) Write the conventional cell representation. (b) Calculate $E_\mathrm{cell}$ using the Nernst equation. (c) Identify the anode and cathode and explain the direction of electron flow.

Solution:

(a)

$$\mathrm{Cu}(s) \mid \mathrm{Cu}^{2+}(0.00100\,\mathrm{mol/dm}^3) \parallel \mathrm{Cu}^{2+}(1.00\,\mathrm{mol/dm}^3) \mid \mathrm{Cu}(s)$$

(b) The overall reaction is:

$$\mathrm{Cu}^{2+}(\text{concentrated}) + \mathrm{Cu}(s) \rightleftharpoons \mathrm{Cu}(s) + \mathrm{Cu}^{2+}(\text{dilute})$$

Using the Nernst equation:

$$E_\mathrm{cell} = E^\circ_\mathrm{cell} - \frac{RT}{nF}\ln Q = 0 - \frac{RT}{2F}\ln\frac{[\mathrm{Cu}^{2+}]_\mathrm{dilute}}{[\mathrm{Cu}^{2+}]_\mathrm{concentrated}}$$ $$E_\mathrm{cell} = -\frac{0.02569}{2}\ln\frac{0.00100}{1.00} = -0.01285 \times \ln(10^{-3}) = -0.01285 \times (-6.908) = +0.0888\,\mathrm{V}$$

(c) The half-cell with the lower concentration ($0.00100\,\mathrm{mol/dm}^3$) is the anode (oxidation: $\mathrm{Cu} \to \mathrm{Cu}^{2+} + 2e^-$). The half-cell with the higher concentration ($1.00\,\mathrm{mol/dm}^3$) is the cathode (reduction: $\mathrm{Cu}^{2+} + 2e^- \to \mathrm{Cu}$). Electrons flow from the dilute side (anode) to the concentrated side (cathode).

Problem 6

In the electrolysis of molten $\mathrm{NaCl}$A current of $2.00\,\mathrm{A}$ is passed for 2.00 hours. Calculate: (a) The mass of sodium produced at the cathode. (b) The volume of chlorine gas produced at the anode at $298\,\mathrm{K}$ and $100\,\mathrm{kPa}$.

Solution:

(a) Cathode: $\mathrm{Na}^+ + e^- \to \mathrm{Na}(l)$. $n = 1$.

$Q = It = 2.00 \times 7200 = 14400\,\mathrm{C}$

$n(e^-) = 14400/96485 = 0.1493\,\mathrm{mol}$

$n(\mathrm{Na}) = 0.1493\,\mathrm{mol}$ (1 electron per Na atom)

$m(\mathrm{Na}) = 0.1493 \times 22.99 = 3.43\,\mathrm{g}$

(b) Anode: $2\mathrm{Cl}^- \to \mathrm{Cl}_2 + 2e^-$. $n = 2$.

$n(\mathrm{Cl}_2) = 0.1493/2 = 0.0747\,\mathrm{mol}$

$V = nRT/p = 0.0747 \times 8.314 \times 298 / (100 \times 10^3) = 1.85 \times 10^{-3}\,\mathrm{m}^3 = 1.85\,\mathrm{dm}^3$

The standard electrode potential of a cell indicates whether a redox reaction is feasible:

$E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode}$

If $E^\circ_\mathrm{cell} > +0.27\,\mathrm{V}$The reaction is considered thermodynamically feasible (proceeds to a significant extent) under standard conditions. If $E^\circ_\mathrm{cell} < +0.27\mathrm{V}$The equilibrium lies to the left and the reaction does not proceed significantly.

Will $\mathrm{Zn}$ displace $\mathrm{Cu}$ from $\mathrm{CuSO}_4$ solution?

Half-equations:

• $\mathrm{Zn}^{2+} + 2e^- \rightleftharpoons \mathrm{Zn}$, $E^\circ = -0.76\,\mathrm{V}$
• $\mathrm{Cu}^{2+} + 2e^- \rightleftharpoons \mathrm{Cu}$, $E^\circ = +0.34\mathrm{V}$

Zinc is the more reactive metal (more negative $E^\circ$), so it will be oxidised (anode). Copper ions will be reduced (cathode).

$E^\circ_\mathrm{cell} = 0.34 - (-0.76) = +1.10\,\mathrm{V}$

Since $E^\circ_\mathrm{cell} > +0.27\,\mathrm{V}$The reaction is feasible:

$\mathrm{Zn}(s) + \mathrm{Cu}^{2+}(aq) \to \mathrm{Zn}^{2+}(aq) + \mathrm{Cu}(s)$

The Nernst equation calculates the cell potential under non-standard conditions:

$E = E^\circ - \frac{RT}{nF}\ln Q$

Where $Q$ is the reaction quotient.

Calculate the cell potential for \mathrm{Zn}|\mathrm{Zn}^{2+}(0.010\,\mathrm{mol\,dm^{-3})||\mathrm{Cu}^{2+}(0.001\,\mathrm{mol\,dm^{-3})|\mathrm{Cu} at $298\,\mathrm{K}$.

$E^\circ_\mathrm{cell} = 0.34 - (-0.76) = 1.10\,\mathrm{V}$

$Q = \frac{[\mathrm{Cu}^{2+}]}{[\mathrm{Zn}^{+}]} = \frac{0.001}{0.010} = 0.10$

$E = 1.10 - \frac{8.314 \times 298}{2 \times 96485}\ln(0.10)$

$= 1.10 - \frac{2478}{192970}\ln(0.10)$

$= 1.10 - 0.01284 \times (-2.303)$

$= 1.10 + 0.0296 = 1.13\,\mathrm{V}$

The cell potential is slightly higher than $E^\circ$ because the lower product concentration ([Cu2+]) drives the reaction further to the right.

Lead-acid accumulator:

• Anode (discharge): $\mathrm{Pb}(s) + \mathrm{SO}_4^{2-}(aq) \rightleftharpoons \mathrm{PbSO}_4(s) + 2e^-$, $E^\circ = -0.36\,\mathrm{V}$
• Cathode (discharge): $\mathrm{PbO}_2(s) + \mathrm{SO}_4^{2-}(aq) + 4\mathrm{H}^+(aq) + 2e^- \rightleftharpoons \mathrm{PbSO}_4(s) + 2\mathrm{H}_2\mathrm{O}(l)$, $E^\circ = +1.69\mathrm{V}$

$E^\circ_\mathrm{cell} = 1.69 - (-0.36) = 2.05\,\mathrm{V}$

During charging, the reactions are reversed.

Hydrogen fuel cell:

• Anode: $2\mathrm{H}_2 + 2\mathrm{H}_2\mathrm{O} \to 4\mathrm{H}^+ + 4e^-$
• Cathode: $\mathrm{O}_2 + 4\mathrm{H}^+ + 4e^- \to 2\mathrm{H}_2\mathrm{O}$

Overall: $2\mathrm{H}_2 + \mathrm{O}_2 \to 2\mathrm{H}_2\mathrm{O}$

$E^\circ_\mathrm{cell} = 1.23\,\mathrm{V}$ (theoretical). Practical voltage is lower due to overpotentials and internal resistance.

Worked Example: How long does it take to deposit $0.500\,\mathrm{g}$ of nickel from $\mathrm{NiSO}_4$ solution using a current of $2.50\,\mathrm{A}$?

$\mathrm{Ni}^{2+} + 2e^- \to \mathrm{Ni}$

$n(\mathrm{Ni}) = \frac{m}{M} = \frac{0.500}{58.69} = 0.00852\,\mathrm{mol}$

$Q = n \times z \times F = 0.00852 \times 2 \times 96485 = 1643\,\mathrm{C}$

$t = \frac{Q}{I} = \frac{1643}{2.50} = 657\,\mathrm{s} = 11.0\,\mathrm{min}$

The electrochemical series arranges species in order of their standard electrode potentials:

Species	$E^\circ$ (V)
$\mathrm{Li}^+/\mathrm{Li}$	$-3.03$
$\mathrm{K}^+/\mathrm{K}$	$-2.93$
$\mathrm{Na}^+/\mathrm{Na}$	$-2.71$
$\mathrm{Mg}^{2+}/\mathrm{Mg}$	$-2.37$
$\mathrm{Al}^{3+}/\mathrm{Al}$	$-1.66$
$\mathrm{Zn}^{2+}/\mathrm{Zn}$	$-0.76$
$\mathrm{Fe}^{2+}/\mathrm{Fe}$	$-0.44$
$\mathrm{Ni}^{2+}/\mathrm{Ni}$	$-0.25$
$\mathrm{Sn}^{2+}/\mathrm{Sn}$	$-0.14$
$\mathrm{Pb}^{2+}/\mathrm{Pb}$	$-0.13$
$2\mathrm{H}^+/\mathrm{H}_2$	$0.00$
$\mathrm{Cu}^{2+}/\mathrm{Cu}$	$+0.34$
$\mathrm{I}_2/\mathrm{I}^-$	$+0.54$
$\mathrm{Ag}^+/\mathrm{Ag}$	$+0.80$
$\mathrm{Br}_2/\mathrm{Br}^-$	$+1.07$
$\mathrm{Cl}_2/\mathrm{Cl}^-$	$+1.36$
$\mathrm{F}_2/\mathrm{F}^-$	$+2.87$

A species will reduce any species below it in the series (more negative $E^\circ$). A species will be oxidised by any species above it (more positive $E^\circ$).

Q1 (5 marks)

A voltaic cell is constructed using the half-cells $\mathrm{Fe}^{2+}|\mathrm{Fe}$ ($E^\circ = -0.44\,\mathrm{V}$) and $\mathrm{Ag}^+|\mathrm{Ag}$ ($E^\circ = +0.80\,\mathrm{V}$).

(a) Write the overall cell equation. (2 marks)

(b) Calculate $E^\circ_\mathrm{cell}$. (1 mark)

(c) Identify the anode and cathode. (2 marks)

Mark Scheme:

(a) $\mathrm{Fe}(s) + 2\mathrm{Ag}^+(aq) \to \mathrm{Fe}^{2+}(aq) + 2\mathrm{Ag}(s)$ (1 mark for correct equation, 1 mark for balancing).

(b) $E^\circ_\mathrm{cell} = 0.80 - (-0.44) = +1.24\,\mathrm{V}$ (1 mark).

(c) Anode = Fe electrode (oxidation occurs; Fe is oxidised to $\mathrm{Fe}^{2+}$) (1 mark). Cathode = Ag electrode (reduction occurs; $\mathrm{Ag}^+$ is reduced to Ag) (1 mark).

Q2 (6 marks)

In the electrolysis of molten $\mathrm{PbBr}_2$ using inert electrodes:

(a) Write the half-equations at each electrode. (2 marks)

(b) Explain why the electrolyte must be molten rather than aqueous. (2 marks)

(c) Calculate the volume of bromine gas produced at the anode when a current of $0.500\,\mathrm{A}$ is passed for 30.0 minutes at $298\,\mathrm{K}$ and $101\,\mathrm{kPa}$. (2 marks)

Mark Scheme:

(a) Cathode: $\mathrm{Pb}^{2+} + 2e^- \to \mathrm{Pb}(l)$ (1 mark). Anode: $2\mathrm{Br}^- \to \mathrm{Br}_2(g) + 2e^-$ (1 mark).

(b) In aqueous solution, water would be preferentially discharged at the cathode ($\mathrm{H}^+$ is reduced before $\mathrm{Pb}^{2+}$) and at the anode ($\mathrm{OH}^-$ is oxidised before $\mathrm{Br}^-$ in dilute solution). Molten $\mathrm{PbBr}_2$ ensures only $\mathrm{Pb}^{2+}$ and $\mathrm{Br}^-$ ions are present (1 mark).

(c) $Q = It = 0.500 \times 1800 = 900\,\mathrm{C}$ (0.5 mark). $n(e^-) = 900/96485 = 0.00933\,\mathrm{mol}$ (0.5 marks). $n(\mathrm{Br}_2) = 0.00933/2 = 0.00467\,\mathrm{mol}$ (0.5 marks). $V = nRT/p = 0.00467 \times 8.314 \times 298 / 101000 = 0.114\,\mathrm{dm}^3 = 114\,\mathrm{cm}^3$ (0.5 marks).

Q3 (4 marks)

State and explain two differences between a primary cell and a secondary cell.

Mark Scheme:

1. A primary cell is not rechargeable; a secondary cell is rechargeable (1 mark). Primary cells produce electricity from irreversible chemical reactions; secondary cells can have the reactions reversed by applying an external voltage (1 mark).

2. Primary cells have a limited lifetime (until the reactants are consumed); secondary cells can be recharged many times (1 mark). Secondary cells are more expensive initially but more cost-effective over their lifetime (1 mark).

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## Cross-References

• Chemistry
• Atomic Structure
• Organic Chemistry